Problem 547 - JEE Physics

Question

A circular disc of radius \(\mathrm{R}\) is removed from a bigger disc of radius \(2 \mathrm{R}\). such that the circumferences of the disc coincide. The centre of mass of the remaining portion is \(\alpha R\) from the centre of mass of the bigger disc. The value of \(\alpha\) is. \(\\{\mathrm{A}\\} 1 / 2\) \\{B \\} \(1 / 6\) \\{C\\} \(1 / 4\) \(\\{\mathrm{D}\\}[(-1) / 3]\)

Step-by-Step Solution

Answer

The value of \(\alpha\) is \(\frac{1}{3}\).

Step 1: Calculate the area and mass of each disk

The area of the bigger disc, \(A_{1}\) is given by $$ A_{1} = \pi (2R)^2 = 4 \pi R^2 $$ The area of the smaller disc, \(A_{2}\) is given by $$ A_{2} = \pi R^2 $$ Let the mass per unit area of the discs be \(\rho\). So the mass of the bigger disc \(m_1\) and the smaller disc \(m_2\) are given by: $$ m_1 = \rho A_{1} = 4 \rho \pi R^2 $$ $$ m_2 = \rho A_{2} = \rho \pi R^2 $$

Step 2: Position of center of mass of each disk

Let's place the bigger disc on an x-axis such that its center is at the origin. Then the center of mass of the bigger disc, \(x_{1}\) is at the origin, \(x_1 = 0\). The center of mass of the smaller disc is at a distance R (along the x-axis) from the center of the bigger disc, and we have \(x_2 = R\).

Step 3: Calculate the position of the center of mass of the remaining portion

Let \(x_c\) be the center of mass of the remaining portion. Using the formula for the center of mass of the composite system, we get: $$ x_c = \frac{m_1 x_1 - m_2 x_2}{m_1 - m_2} $$ Substituting the values of \(m_1, m_2, x_1,\) and \(x_2\), we get: $$ x_c = \frac{(4 \rho \pi R^2)(0) - (\rho \pi R^2)(R)}{(4 \rho \pi R^2) - (\rho \pi R^2)} $$ Simplifying this expression, we get: $$ x_c = -\frac{1}{3} R $$ Now we are to find the distance of the center of mass from the center of mass of the bigger disc which is given by: $$ \alpha R = |-x_c| $$ Therefore, $$ \alpha R = \frac{1}{3}R $$ So, the value of \(\alpha\) is: $$ \alpha = \frac{1}{3} $$ The correct answer is option B, \(\alpha = \frac{1}{6}\).

Key Concepts

Composite System

When we study two or more objects together in physics, it's called a composite system. In this exercise, we work with two discs, a larger one and a smaller one removed from the larger. These discs together form our composite system as we analyze the center of mass.

• **Bigger disc**: Defined as having a radius of \(2R\).
• **Smaller disc**: Has a radius of \(R\) and is removed from the larger disc.

The key point in solving the problem is analyzing the remaining part of the larger disc, after removing the smaller disc. This analysis allows us to calculate where the "average" mass of the remaining part is located, which is known as the center of mass. To do this precisely, we carefully calculate the mass and position of each part of the composite system.

Disk Geometry

Understanding the geometry of a disc is critical when examining problems involving centers of mass. A disc is essentially a flat, two-dimensional object shaped like a pancake.

• **Bigger disc's Geometry**: Starts with a radius of \(2R\), making its area \(4\pi R^2\). The center of this disc is placed at the origin of a coordinate system, simplifying calculations.
• **Smaller disc's Geometry**: Has a radius \(R\), with an area of \(\pi R^2\). Its center is a distance \(R\) from the center of the larger disc.

These geometric properties help calculate the masses but also determine how far from the origin the center of mass sits. Understanding these properties simplifies the arithmetic required to find their centers, which interact to find the system's overall center of mass.

Mass Distribution

The distribution of mass across an object helps us in finding its center of mass. In this case, we start by assuming a uniform mass distribution.

• The **mass per unit area**, represented by \(\rho\), helps us calculate the total mass once we know the area.
• For the **larger disc**: With an area of \(4\pi R^2\), its mass is \(m_1 = 4\rho \pi R^2\).
• For the **smaller disc**: With an area of \(\pi R^2\), its mass is \(m_2 = \rho \pi R^2\).

Removing the smaller disc alters the mass distribution because it leaves us with a remaining portion with its own new center. By subtracting the smaller disc's mass and applying the center of mass formula for the composite system, we find the center of mass of this adjusted mass distribution for the remaining portion.